At some point I thought this is due to the limited information in such a calculation, it is binomial and one simulation gives one bit of information. And it could be more simple. If the first random number is known, say y, then all second random numbers over sqrt(1-y2) give distance larger than 1, while the remainder gives distance less than 1. Thus should pi be equal to the mean of random numbers transformed like sqrt(1-y2)?pin <- function(N=1000) { 4*sum(sqrt(1-runif(N)^2))/N}summary(sapply(1:1000,function(x) pin(10000))) Min. 1st Qu. Median Mean 3rd Qu. Max. 3.113 3.135 3.142 3.141 3.147 3.171
These numbers are closer, but there are additional calculations. Hence the number of simulations should be adapted to reflect the work done. Luckily we have microbenchmark() to calibrate this. After a bit of experimenting, these are the number of simulations giving roughly equivalent computation times.microbenchmark(pi2d(10000),pi3d(6666),pin(22000))

Unit: milliseconds

expr min lq mean median uq max neval

pi2d(10000) 2.419106 2.436333 2.630764 2.458325 2.500477 5.253860 100

pi3d(6666) 2.361928 2.382820 2.557150 2.418006 2.467855 4.970898 100

pin(22000) 2.448429 2.468954 2.555823 2.485815 2.517703 5.023678 100

As can be seen, the third calculation actually has more simulations. Hence it is much more efficient to obtain the estimate.

Sunday, February 14, 2016

Since I read documents with Clopper-Pearson a number of times the last weeks, I thought it a good idea to play around with confidence intervals for proportions a bit; to examine how intervals differ between various approaches. From a frequentist side Clopper-Pearson, which is described as the frequentist's gold standard and secondly the easy way normal approximation. From the Bayesian side, binomial with beta Beta prior. Obviously, the intervals have completely different interpretation in the frequentist and Bayesian framework, but that is a different discussion. There will be no data in this analysis, I am just making intervals based on possible results

Code

There are many ways to set this up. I wanted some plots. My first approach; given an observed proportion of 'correct', how does the total of trials change the intervals? The second approach; given that a certain number of trials is done, how do the intervals change as the number correct changes?

Since I want to repeat many of these calculations, I first made some supporting functions. This is because I am trying to write more clear code, where as much as possible code is not repeated but rather delegated to some sort of function. That may not result in the shortest or fastest code, but at this point neither is required.

Intervals

The first functions create the intervals from n (observed) and N (total). Clopper-Pearson is extracted from binom.test(). Normal approximation is based on an internet example. Beta-Binomial has three functions, one for the actual work, two to set up the desired priors and adapt the naming. A final function calls all these.

clopper.pearson <- function(n,N,conf.level=0.95) {

limits <- as.numeric(binom.test(n,N,conf.level=conf.level)$conf.int)

names(limits) <- c('cp_low','cp_high')

limits

}

binom.norm.app <- function(n,N,conf.level=0.95) {

# based on http://www.r-tutor.com/elementary-statistics/interval-estimation/interval-estimate-population-proportion

phat <- n/N

shat <- sqrt(phat*(1-phat)/N)

limit <- (1-conf.level)/2

zlim <- qnorm( c(limit,1-limit))*shat

limits <- phat+zlim

names(limits) <- c('na_low','na_high')

limits

}

beta.binomial <- function(n,N,conf.level=0.95,prior=c(1,1)){

limit <- (1-conf.level)/2

limits <- qbeta(c(limit,1-limit),n+prior[1],N-n+prior[2])

names(limits) <- c('bb_low','bb_high')

limits

}

beta.binomial11 <- function(n,N,conf.level=0.95) {

limits <- beta.binomial(n,N,conf.level=conf.level,prior=c(1,1))

names(limits) <- c('bb11_low','bb11_high')

limits

}

beta.binomial.5.5 <- function(n,N,conf.level=0.95) {

limits <- beta.binomial(n,N,conf.level=conf.level,prior=c(0.5,0.5))

names(limits) <- c('bb.5_low','bb.5_high')

limits

}

all.intervals <- function(n,N,conf.level=0.95) {

c(n=n,N=N,conf.level=conf.level,

clopper.pearson(n,N,conf.level=conf.level),

binom.norm.app(n,N,conf.level=conf.level),

beta.binomial11(n,N,conf.level=conf.level),

beta.binomial.5.5(n,N,conf.level=conf.level))

}

Post processing

Just doing an sapply() on all.intervals() gives a matrix. It needs to be processed a bit to get a nice data.frame which ggplot likes. Hence a function in which it is transposed, reshaped and names of the intervals are split. Naming is adapted for display purposes.

Results

Results for a proportion correct

The codes are variations on this example for 50% correct. As most of the work is done in the supporting functions, there is no need to repeat the code:

have1outN <- sapply(1:20,function(x) all.intervals(1*x,2*x))

have1outN <- postprocessing(have1outN)

ggplot(have1outN,aes(x=limit,y=N,col=Method,l=direction)) +

geom_path() +

xlim(c(min(0,have1outN$limit),max(1,have1outN$limit))) +

ggtitle('Interval at 1/2 correct') +

theme(legend.position="bottom") +

guides(col=guide_legend(ncol=2))

It seems that especially at lower N the Normal approximation is not advisable. Having an interval stick outside the range 0-1 is obviously a dead giveaway that something is not correct. But even if that does not happen, the lines are pretty far of the remainder of the methods. The difference between the two Beta Binomials is surprisingly small and only visible when very few observations are made. Clopper-Pearson seems to give slightly wider intervals than Beta Binomial.

Results for a fixed N

Again, the code is variations on a theme, with the work being done by the supporting functions.

Again the normal approximation is the odd out. It also seems to degenerate at n=0 and n=N. Other than that the choice of prior in the Beta Binomial is more expressed that the previous plots.

Tuesday, February 2, 2016

A couple of years I have made plots of unemployment and its change over the years. At first this was a bigger and complex piece of code. As things have progressed, the code can now become pretty concise. There are just plenty of packages to do the heavy lifting. So, this year I tried to make the code easy to read and reasonably documented.

Data

Data is from Eurostat. Since we have the joy of the Eurostat package, suffice to say this is dataset une_rt_m. Since the get_eurostat function gave me codes for things such as country and gender, the first step is to use a dictionary to decode. Subsequently, the country names are a bit sanitized and data is selected.library(eurostat)library(ggplot2)library(KernSmooth)library(plyr)library(dplyr)library(scales) # to access breaks/formatting functions

Plots

To make plots I want to have smoothed data. Ggplot will do this, but it is my preference to have the same smoothing for all curves, hence it is done before entering ggplot. There are a bit many countries, hence the number is reduced to 36, which are displayed in three plots of 3*4, for countries with low, middle and high maximum unemployment respectively. Two smoothers are applied, once for the smoothed data, the second for its first derivative. The derivative has forced more smooth, to avoid extreme fluctuation.

# add 3 categories for the 3 3*4 displays

r3 <- aggregate(r2$values,by=list(geo=r2$geo),FUN=max,na.rm=TRUE) %>%

mutate(.,class=cut(x,quantile(x,seq(0,3)/3),

include.lowest=TRUE,

labels=c('low','middle','high'))) %>%

select(.,-x) %>% # maxima not needed any more

right_join(.,r2)

#locpoly to make smooth same for all countries

Perc <- ddply(.data=r3,.variables=.(age,geo),

function(piece,...) {

piece <- piece[!is.na(piece$values),]

lp <- locpoly(x=as.numeric(piece$time),y=piece$values,

drv=0,bandwidth=90)

sdf <- data.frame(Date=as.Date(lp$x,origin='1970-01-01'),

sPerc=lp$y,

age=piece$age[1],

geo=piece$geo[1],

country=piece$country[1],

class=piece$class[1])}

,.inform=FALSE

)

# locpoly for deriviative too

dPerc <- ddply(.data=r3,.variables=.(age,geo),

function(piece,...) {

piece <- piece[!is.na(piece$values),]

lp <- locpoly(x=as.numeric(piece$time),y=piece$values,

drv=1,bandwidth=365/2)

sdf <- data.frame(Date=as.Date(lp$x,origin='1970-01-01'),

dPerc=lp$y,

age=piece$age[1],

geo=piece$geo[1],

country=piece$country[1],

class=piece$class[1])}

,.inform=FALSE

)

The plots are processed by subsection.

for (i in c('low','middle','high')) {

png(paste(i,'.png',sep=''))

g <- filter(Perc,class==i) %>%

ggplot(.,

aes(x=Date,y=sPerc,colour=age)) +

facet_wrap( ~ country, drop=TRUE) +

geom_line() +

theme(legend.position = "bottom")+

ylab('% Unemployment') + xlab('Year') +

scale_x_date(breaks = date_breaks("5 years"),

labels = date_format("%y"))

print(g)

dev.off()

}

for (i in c('low','middle','high')) {

png(paste('d',i,'.png',sep=''))

g <- filter(dPerc,class==i) %>%

ggplot(.,

aes(x=Date,y=dPerc,colour=age)) +

facet_wrap( ~ country, drop=TRUE) +

geom_line() +

theme(legend.position = "bottom")+

ylab('Change in % Unemployment') + xlab('Year')+

scale_x_date(breaks = date_breaks("5 years"),

labels = date_format("%y"))

print(g)

dev.off()

}

Results

In general, things are improving, which is good news, though there is still ways to go. As always, Eurostat has a nice document are certainly more knowledgeable than me on this topic.

Average unemployment

First derivative

Sunday, January 17, 2016

I was browsing Davies Design and Analysis of Industrial Experiments (second edition, 1967). Published by for ICI in times when industry did that kind of thing. It is quite an applied book. On page 107 there is an example where the variance of a process is estimated.

Data

Data is from nine batches from which three samples were selected (A, B and C) and each a duplicate measurement. I am not sure about copyright of these data, so I will not reprint the data here. The problem is to determine the measurement ans sampling error in a chemical process.ggplot(r4,aes(x=Sample,y=x))+ geom_point()+ facet_wrap(~ batch )

Analysis

At the time of writing the book, the only approach was to do a classical ANOVA and calculate the estimates from there.aov(x~ batch + batch:Sample,data=r4) %>% anovaAnalysis of Variance TableResponse: x Df Sum Sq Mean Sq F value Pr(>F) batch 8 792.88 99.110 132.6710 < 2e-16 ***batch:Sample 18 25.30 1.406 1.8818 0.06675 . Residuals 27 20.17 0.747 ---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1In this case the residual variation is 0.75. The batch:Sample variation estimates is, due to the design, twice the sapling variation plus residual variation. Hence it is estimated as 0.33. How lucky we are to have tools (lme4) which can do this estimate directly. In this case, as it was a well designed experiment, these estimates are the same as from the ANOVA. l1 <- lmer(x ~1+ (1 | batch) + (1|batch:Sample) ,data=r4 )

Sunday, January 3, 2016

I ran across this post containing displays on who works from home. I must say it looks great and is interactive but it did not help me understand the data. So I created this post to display the same data with a boring plot which might help me. For those really interested in this topic, census.gov created a .pdf which contains a full report with much more information than here.

Data

Data is from census.gov. I have taken the first spreadsheet. It is one of those spreadsheets with counts and percentages and empty lines to display categories. Very nice to check some numbers, horrible to process. So, a bit of code to extract the numbers.library(gdata)r1 <- read.xls('2010_Table_1.xls',stringsAsFactors=FALSE)# throw out percentagesr2 <- r1[,r1[4,]!='Percent']# put all column names in one rowr2$X.6[2] <- r2$X.6[3]r2$X.8[2] <- r2$X.8[3]# select part with datar3 <- r2[2:61,c(1,3,5,6)]names(r3)[1] <- r3[1,1]r4 <-r3[c(-1:-3),]#eliminate one row with mean income. r4 <- r4[-grep('$',r4[,2],fixed=TRUE),]#reshape in long formr5 <- reshape(r4, varying=list(names(r4)[-1]), v.names='count', direction='long', idvar='Characteristic', timevar='class', times=r3[1,2:4])row.names(r5) <- 1:nrow(r5)# remove ',' from numbers and make numerical values. # units are in 1000, so update that toor5$count <- as.numeric(gsub(',','',r5$count))*1000# clean up numbers used for footnotesr5$class <- gsub('(1|2|3)','',r5$class)#some upfront '.' removed.r5$Characteristic <- gsub('^\\.+','',r5$Characteristic)# create a factorr5$Characteristic <- factor(r5$Characteristic, levels=rev(r5$Characteristic[r5$class=='Home Workers']))# and create a higher level factorr5$Mchar=r5$Characteristicfor (i in 1:nrow(r5)) r5$Mchar[i] <- if(is.na(r5$count[i]) | r5$Mchar[i]=='Total') r5$Mchar[i] else r5$Mchar[i-1]

Wiekvoet

Wiekvoet is about R, JAGS, STAN, and any data I have interest in. Topics range from sensometrics, statistics, chemometrics and biostatistics. For comments or suggestions please email me at wiekvoet at xs4all dot nl.